Question

fill in the missing words
if the following reaction is balanced correctly $c_{4}h_{10}+o_{2}
ightarrow co_{2}+h_{2}o$ (and all coefficients are whole numbers), the coefficient in front of $o_{2}$ is
, and the coefficient in front of $h_{2}o$ is .

Explanation:

Step1: Balance carbon atoms

There are 4 carbon atoms in $C_{4}H_{10}$. So, the coefficient of $CO_{2}$ should be 4 to balance carbon. The reaction becomes $C_{4}H_{10}+O_{2}
ightarrow4CO_{2}+H_{2}O$.

Step2: Balance hydrogen atoms

There are 10 hydrogen atoms in $C_{4}H_{10}$. So, the coefficient of $H_{2}O$ should be 5 to balance hydrogen. The reaction is now $C_{4}H_{10}+O_{2}
ightarrow4CO_{2}+5H_{2}O$.

Step3: Balance oxygen atoms

On the right - hand side, there are $4\times2 + 5\times1=8 + 5 = 13$ oxygen atoms. So, the coefficient of $O_{2}$ should be $\frac{13}{2}$. But we need whole - number coefficients. Multiply the entire equation by 2 to get $2C_{4}H_{10}+13O_{2}
ightarrow8CO_{2}+10H_{2}O$.

Answer:

The coefficient in front of $O_{2}$ is 13, and the coefficient in front of $H_{2}O$ is 10.